Question: If $x \oplus y = (8-x)(y)$ and $x \veebar y = 3x+y$, find $0 \veebar (3 \oplus -2)$.
Answer: First, find $3 \oplus -2$ $ 3 \oplus -2 = (8-3)(-2)$ $ \hphantom{3 \oplus -2} = -10$ Now, find $0 \veebar -10$ $ 0 \veebar -10 = (3)(0)-10$ $ \hphantom{0 \veebar -10} = -10$.